Problem: Suppose we have a vector field $f(x, y) = (x, \cos(y))$ and a curve $C$ that is parameterized by $\alpha(t) = \left( t^2, t \right)$ for $0 < t < 2$. What is the line integral of $f$ along $C$ ? $ \int_C f \cdot d\alpha = $
Given a vector field $f$, a parameterization $\alpha$, and bounds $t_0$ and $t_1$, we can calculate the line integral as follows: $ \int_C f \cdot d\alpha = \int_{t_1}^{t_2} f(\alpha(t)) \cdot \alpha'(t) \, dt$ Here, $f(x, y) = (x, \cos(y))$ and $\alpha(t) = (t^2, t)$. $\begin{aligned} &f(\alpha(t)) = \left( t^2, \cos(t) \right) \\ \\ &\alpha'(t) = (2t, 1) \end{aligned}$ Now we can rewrite our line integral as a single-variable integral. $ \int_C f \cdot d\alpha = \int_0^2 \left( t^2, \cos(t) \right) \cdot (2t, 1) \, dt$ Let's solve the integral. $\begin{aligned} \int_0^2 \left( t^2, \cos(t) \right) \cdot (2t, 1) \, dt &= \int_0^2 2t^3 + \cos(t) \, dt \\ \\ &= \left[ \dfrac{t^4}{2} + \sin(t) \right]_0^2 \\ \\ &= 8 + \sin(2) \end{aligned}$ In conclusion, the line integral $ \int_C f \cdot d\alpha = 8 + \sin(2)$.